# Question Video: Calculating the Power of a Matrix Mathematics • 10th Grade

Given that 𝐴 = [2, 0 and 2, 0], which of the following is 𝐴⁶⁰? [A] 2⁶⁰[2, 0 and 2, 0] [B] 2⁶⁰[1, 0 and 1, 0] [C] [1, 0 and 1, 0] [D] 2[1, 0 and 1, 0] [E] 2⁶⁰[1, 1 and 1, 1]

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### Video Transcript

Given that 𝐴 is the two-by-two matrix two, zero, two, zero, which of the following is 𝐴 to the 60th power? Is it option (A) two to the 60th power multiplied by the two-by-two matrix two, zero, two, zero? Is it option (B) two to the 60th power multiplied by the two-by-two matrix one, zero, one, zero? Is it option (C) the two-by-two matrix one, zero, one, zero? Option (D) two times the two-by-two matrix one, zero, one, zero. Or is it option (E) two to the 60th power multiplied by the two-by-two matrix one, one, one, one?

In this question, we’re given a two-by-two matrix 𝐴. And we need to determine which of five given options is a representation of 𝐴 to the 60th power. To answer this question, let’s start by recalling what it means to raise a matrix to an exponent. We recall if 𝐴 is a square matrix and the exponent is a positive integer, then this means we multiply the matrix by itself so that it appears the exponent number of times in the product. In particular, this tells us 𝐴 to the 60th power is 𝐴 multiplied by itself, where 𝐴 appears 60 times in the product.

And of course we’re not going to be able to evaluate this product directly. Since 𝐴 is a two-by-two matrix, this means we’re multiplying 60 two-by-two matrices together. Instead, we can simplify this expression slightly by using the properties of associativity of matrix multiplication, which of course tells us we can evaluate this product in any order. In particular, we can simplify 𝐴 times 𝐴 to be 𝐴 squared. So let’s find an expression for 𝐴 squared. Remember, that’s 𝐴 times 𝐴, which is the two-by-two matrix two, zero, two, zero multiplied by itself.

To find the product of two matrices, we need to find the sum of the products of the corresponding elements of the rows of the first matrix with the columns of the second matrix. For example, the entry in row one, column one of matrix 𝐴 squared will be two times two plus zero times two, which we can calculate is equal to four. We can follow the same method to find the element in row one, column two. It’s two times zero plus zero times zero. And both of these terms have a factor of zero. So this evaluates to give us zero. We can continue this process to find the element in row two, column one. It’s two times two plus zero times two, which we can evaluate is equal to four. Finally, we can find the element in row two, column two. It’s equal to two times zero plus zero times zero, which we can calculate is zero.

Therefore, we’ve shown 𝐴 squared is the two-by-two matrix four, zero, four, zero. And this is a really useful result. We can see this is a scalar multiple of matrix 𝐴. We take out the scalar of two from this matrix. This gives us two multiplied by the two-by-two matrix two, zero, two, zero, which is just the matrix 𝐴. So 𝐴 squared is equal to two times 𝐴.

And we can now use this equation to evaluate 𝐴 raised to any positive integer exponent. To see how we can do this, let’s use this to evaluate 𝐴 cubed. First, we know that 𝐴 cubed is equal to 𝐴 times 𝐴 times 𝐴, which we can rewrite as 𝐴 squared times 𝐴. We can then replace 𝐴 squared with two multiplied by 𝐴. This gives us two 𝐴 multiplied by 𝐴, which we can simplify to give us two 𝐴 squared. But remember, 𝐴 squared is equal to two 𝐴. Therefore, this is equal to two multiplied by two 𝐴. And two times two is equal to four. Therefore, 𝐴 cubed is equal to four times 𝐴.

Therefore, we’ve shown 𝐴 squared is equal to two 𝐴 and 𝐴 cubed is equal to four 𝐴. And we can start to notice a pattern. Whenever we multiply by an extra factor of 𝐴, we seem to multiply 𝐴 by an extra factor of two. And we can prove that this is true. Let’s multiply both sides of our equation through on the right by matrix 𝐴. This gives us 𝐴 cubed multiplied by 𝐴 is equal to four times 𝐴 times 𝐴.

First, we simplify 𝐴 cubed times 𝐴 is 𝐴 to the fourth power. Next, 𝐴 times 𝐴 is equal to 𝐴 squared. And we’ve already shown 𝐴 squared is equal to two 𝐴. Therefore, 𝐴 to the fourth power is equal to four multiplied by two 𝐴, which is eight 𝐴. Every time we multiply on the right by 𝐴, we get an extra factor of two.

Let’s now use this property to find an expression for 𝐴 to the 60th power. To do this, we need to multiply 𝐴 to the fourth power by 𝐴 56 times. And each time we multiply by 𝐴, we get an extra factor of two. We do this 56 times, so we get 56 factors of two. That’s two to the power of 56. Therefore, 𝐴 to the 60th power is eight multiplied by two to the power of 56 𝐴.

However, this is not one of our options, so we need to simplify further. We’ll do this by first noting that eight is two cubed. And two cubed multiplied by two to the power of 56 is two to the power of 59. And this is still not any of our options. Instead, we need to notice matrix 𝐴 also has a factor of two. Matrix 𝐴 is two times the two-by-two matrix one, zero, one, zero. So rewriting 𝐴 in this manner and simplifying, we get two to the 60th power times the two-by-two matrix one, zero, one, zero, which we can see is option (B).

Therefore, we were able to show if 𝐴 is the two-by-two matrix two, zero, two, zero, then 𝐴 to the 60th power is equal to two to the 60th power multiplied by the two-by-two matrix one, zero, one, zero, which is option (B).